Use synthetic division to solve problems Evaluate polynomial functions using the remainder theorem

1. Synthetic Division
Chapter 3 Polynomial and Rational Functions

2. Concepts and Objectives
 Synthetic Division
 Review performing synthetic division
 Evaluate polynomial functions using the remainder
theorem

3. Dividing Polynomials
 Back in section 0.3, we found the quotient of two
polynomials by using something like long division.
 More formally, we can state that:
Let fx and gx be polynomials with gx of degree one
or more, but of lower degree than fx. There exist
unique polynomials qx and rx such that
where either rx = 0 or the degree of rx is less than the
degree of gx.
        f x g x q x r x

4. Dividing Polynomials (cont.)
 For example, could be evaluated as
or
 

3 2
2
3 2 150
4
x x
x
   2 3 2
3
4 3 2 0 150
x
x x x x
  3 2
3 0 12x x x
  2
2 12 150x x
2
 2
2 0 8x x
12 158x

 
2
12 158
3 2
4
x
x
x

5. Dividing Polynomials (cont.)
 Using the division algorithm, this means that
        3 2 2
3 2 150 4 3 2 12 158x x x x x
 f x  g x  q x  r x
Dividend = Divisor • Quotient + Remainder

6. Synthetic Division
 A shortcut method of performing long division with
certain polynomials, called synthetic division, is used
only when a polynomial is divided by a binomial of the
form x – k, where the coefficient of x is 1.
 To use synthetic division:
The numbers on the bottom are the coefficients of the
quotient.
1 1 0...n nk a a a a

   


1
1 1 0...n n
n na x a x a x a
x k
an
kan
 1n na ka …

7. Synthetic Division (cont.)
 Example: Use synthetic division to divide
  

3 2
4 15 11 10
3
x x x
x

8. Synthetic Division (cont.)
 Example: Use synthetic division to divide
  

3 2
4 15 11 10
3
x x x
x
 3 4 15 11 10

9. Synthetic Division (cont.)
 Example: Use synthetic division to divide
  

3 2
4 15 11 10
3
x x x
x
 3 4 15 11 10
4
12
–3

10. Synthetic Division (cont.)
 Example: Use synthetic division to divide
  

3 2
4 15 11 10
3
x x x
x
 3 4 15 11 10
4
12
–3
–9
2

11. Synthetic Division (cont.)
 Example: Use synthetic division to divide
  

3 2
4 15 11 10
3
x x x
x
 3 4 15 11 10
4
12
–3
–9
2
6
–4

12. Synthetic Division (cont.)
 Example: Use synthetic division to divide
  

3 2
4 15 11 10
3
x x x
x
 3 4 15 11 10
4
12
–3
–9
2
6
–4

  

2 4
4 3 2
3
x x
x
Notice that the
exponent is 1 less
than the numerator.

13. Synthetic Division (cont.)
 Example: Use synthetic division to divide
   

4 3
3 15 50 25
4
x x x
x

14. Synthetic Division (cont.)
 Example: Use synthetic division to divide
   

4 3
3 15 50 25
4
x x x
x
 4 3 15 500 25
Notice the place-
holder for the
missing x2 term.

15. Synthetic Division (cont.)
 Example: Use synthetic division to divide
   

4 3
3 15 50 25
4
x x x
x
 4 3 15 500 25
3
–12
3
12
12
48
2
–8
17
    

3 2 17
3 3 12 2
4
x x x
x
Notice the place-
holder for the
missing x2 term.

16. Synthetic Division (cont.)
 Example: Use synthetic division to divide
   

4 3 2
5 4 3 9
3
x x x x
x

17. Synthetic Division (cont.)
 Example: Use synthetic division to divide
   

4 3 2
5 4 3 9
3
x x x x
x
 3 1 5 4 3 9
1
–3
2
–6
–2
6
3
–9
0
  3 2
2 2 3x x x

18. Remainder Theorem
 The remainder theorem:
 Example: Let . Find f–3.
If the polynomial fx is divided by x – k, then the
remainder is equal to fk.
     4 2
3 4 5f x x x x

19. Remainder Theorem
 The remainder theorem:
 Example: Let . Find f–3.
If the polynomial fx is divided by x – k, then the
remainder is equal to fk.
     4 2
3 4 5f x x x x
   3 1 0 3 4 5
–1
3
3
–9
–6
18
14
–42
–47
   3 47f

20. Potential Zeros
 A zero of a polynomial function f is a number k such that
fk = 0. The real number zeros are the x-intercepts of
the graph of the function.
 The remainder theorem gives us a quick way to decide if
a number k is a zero of a polynomial function defined by
fx. Use synthetic division to find fk ; if the remainder
is 0, then fk = 0 and k is a zero of fx .

21. Potential Zeros (cont.)
 Example: Decide whether the given number k is a zero
of fx:
       4 3 2
4 14 36 45; 3f x x x x x k

22. Potential Zeros (cont.)
 Example: Decide whether the given number k is a zero
of fx:
Since the remainder is zero, –3 is a zero of the function.
       4 3 2
4 14 36 45; 3f x x x x x k
  3 1 4 14 36 45
1
–3
–7
21
7
–21
15
–45
0

23. Classwork
 3.2 Assignment (College Algebra)
 Page 326: 2-14 (even), page 317: 60-64 (even),
page 313: 38, 40
 3.2 Classwork Check
 Quiz 3.1b